Where do mass polarization terms come from in many-body Hamiltonian? Why are they sometimes omitted? This question is about the Hamiltonian for more than one particle (non-relativistic).
Griffiths (Introduction to Quantum Mechanics, 2e) seems to imply that it is $\displaystyle H=-\frac{\hbar^2}{2}\left(\sum_{n=1}^N\frac{1}{m_n}\nabla_{\mathbf{r}_n}^2\right)+V(\mathbf{r}_1,\dots,\mathbf{r}_N,t)$, but wikipedia isn't so clear: https://en.wikipedia.org/wiki/Hamiltonian_%28quantum_mechanics%29#Many_particles. 
Initially the article quotes that formula, but it quickly gets confusing:  

However, complications can arise in the many-body problem. Since the potential energy depends on the spatial arrangement of the particles, the kinetic energy will also depend on the spatial configuration to conserve energy. The motion due to any one particle will vary due to the motion of all the other particles in the system. For this reason cross terms for kinetic energy may appear in the Hamiltonian; a mix of the gradients for two particles:
  $-\frac{\hbar^2}{2M}\nabla_i\cdot\nabla_j$
  where M denotes the mass of the collection of particles resulting in this extra kinetic energy. Terms of this form are known as mass polarization terms, and appear in the Hamiltonian of many electron atoms (see below).

Unfortunately, the author wrote nothing below which might explain where the mass polarization terms come from.
Could I get some mathematical background, maybe a derivation, of why these terms are present in the Hamiltonian, and why they are sometimes omitted/forgotten?
 A: I've never heard of this before and it seems rather weird, but I'll have a guess that seems like the only reasonable thing that could be going on. The kinetic energy of one particle definitely does not depend (fundamentally) on what another particle is doing. It only indirectly depends on the other particles through interactions (i.e. potentials or gauge fields). The author must have in mind some sort of Born-Oppenheimer-like approximation where some subset of interactions are taken care of or fixed before hand, leaving correlations between the remaining degrees of freedom. So here is my guess about what is happening.
If you have two equal mass particles with momenta $\vec{p}_1,\vec{p}_2$ then the total energy is
$$ H = \frac{\vec{p}_1^2}{2m} + \frac{\vec{p}_2^2}{2m} + V(\vec{r}_1,\vec{r}_2).$$
Now introduce the total $\vec{P}=\vec{p}_1+\vec{p}_2$ and relative momentum $\vec{p}=(\vec{p}_1-\vec{p}_2)/2$ then the Hamiltonian becomes
$$ H = \frac{\vec{P}^2}{2M} + \frac{\vec{p}^2}{2\mu} + V(\vec{r}_1,\vec{r}_2),$$
where $M=2m$ is the total mass and $\mu=m/2$ the reduced mass. Now if you neglect the relative momentum because the two particles are rigidly bound and move together always, then you get
$$ H \approx \frac{\vec{P}^2}{2M} + V(\vec{r}_1,\vec{r}_2). $$
Finally substituting back in you get a cross term $\propto \vec{p}_1\cdot\vec{p}_2$ of the kind mentioned in the article, and the mass $M$ indeed refers to the mass of a conglomerate of several particles. You can also do this if the potential is such that you can separate the centre of mass and relative motion problems and solve the relative motion problem on its own. And this idea should straightforwardly extend to the $N$ particle case. I leave that to you.
A: The mass polarization term, arises when we consider the motion of the nucleus as well(or in other words, give it a finite amount of mass). The kinetic energy of an atom with a nucleus of mass $M$ and charge $Ze$ and $N$ electrons of mass $m$ and charge $-e$ can be written as:
$$T=-\frac{\hbar^2}{2M}\nabla_{R_0}^2+\sum_{i=1}^N \left(-\frac{\hbar^2}{2m}\nabla_{R_i}^2\right) $$
where we have denoted the coordinates of the nucleus with $R_0$ and the electrons with $R_i$.
Now, moving to the center of mass frame; we define new coordinates $(\vec R,\vec r_1,\vec r_2,\dots,\vec r_N)$, where:
$$\vec R=\frac{1}{M+Nm}(M\vec{R_0}+m\vec{R_1}+\dots+m\vec {R_N})$$
is the center of mass, and
$$\vec{r_i}=\vec{R_i}-\vec{R_0}$$
are the relative coordinates. From the above two equations, one can easily verify that:
$$\nabla_{R_0}=\frac M{M+Nm}\nabla_R-\sum_{i=1}^N \nabla_{r_i}$$
$$\nabla_{R_i}= \frac{m}{M+Nm}\nabla_R+\nabla_{r_i} $$
Hence, we will have:
$$\nabla_{R_0}^2=\left(\frac{M}{M+Nm}\right)^2\nabla_R^2-\frac{2M}{M+Nm}\sum_{i=1}^N \nabla_R . \nabla_{r_i}+\left(\sum_{i=0}^N\nabla_{r_i}\right)^2$$
$$\nabla_{R_i}^2=\left(\frac{m}{M+Nm}\right)^2\nabla_R^2+\frac{2m}{M+Nm} \nabla_R . \nabla_{r_i}+\nabla_{r_i}^2$$
Now, substituting these back into the initial equation we get:
\begin{align}
T=&-\frac{\hbar^2}{2M}\left(\left(\frac{M}{M+Nm}\right)^2\nabla_R^2-\frac{2M}{M+Nm}\sum_{i=1}^N \nabla_R \cdot \nabla_{r_i}+
\left(\sum_{i=0}^N\nabla_{r_i}\right)^2\right)+\\
&\quad\sum_{i=1}^N\left(-\frac{\hbar^2}{2m}\left(\left(\frac{m}{M+Nm}\right)^2\nabla_R^2+\frac{2m}{M+Nm} \nabla_R . \nabla_{r_i}+\nabla_{r_i}^2\right)\right)\\
=&-\frac{\hbar^2}{2(M+Nm)^2}(M+Nm)\nabla_R^2-\frac{\hbar^2}{2M}\sum_{i,j=1}^n\nabla_{r_i}\cdot\nabla_{r_j}-\frac{\hbar^2}{2m}\sum_{i=1}^n\nabla_{r_i}^2\\ 
=& 
-\frac{\hbar^2}{2(M_{tot})}\nabla_R^2-\frac{\hbar^2}{2m}\sum_{i=1}^n\nabla_{r_i}^2 -\frac{\hbar^2}{2M}\sum_{i,j=1}^n\nabla_{r_i}\cdot\nabla_{r_j}
\end{align}
where the last is the mass polarization term.
